All categories

4 years ago

What can you use to stick blocks of snow together answer sheet?

Physics

1 answer:

solong [7]4 years ago

5 0

Answer: you can use salt and a bit of heat the

Explanation:put salt on the place you want it to stick then heat it up a littel and it should freeze together

You might be interested in

What is the net force acting on a golf car traveling at a constant speed of 5 mph?

MAVERICK [17]

Technically, we can't answer the question with the given information.
Even though the cart's speed is constant, it may be turning an corner,
or driving along a curved path. If it's not driving in a straight, then there
must be some force acting on it.

If the cart IS driving in a straight line, AND its speed is constant, then
there can't be any net force acting on it. In that case, the correct choice
is ' B '. (I'm sure this is right IF the cart is driving in a straight line.)

3 0

3 years ago

The astrometric technique of planet detection works best for

denpristay [2]

The astrometric technique of planet detection works best for massive planets around nearby stars.

6 0

3 years ago

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

3 0

3 years ago

The classic Millikan oil-drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oi

Sati [7]

Answer:

$1.56\times 10^5 N/C$

Explanation:

The electric field that will balance the weight of the oil drop can be calculated using the following:

electric force, F = e E ( where, e is the charge of an electron and E is the electric field)

weight, W = 2.5 ×10⁻¹⁴ N

e E = W

$E =\frac{W}{e}$

Substitute the values:

$E =\frac{ 2.5 \times 10^{-14} N}{1.6\times10^{-19}C}= 1.56\times 10^5 N/C$

6 0

3 years ago

Read 2 more answers

A proton is traveling horizontally to the right at 1.8 × 106 m/s. (a) Find the magnitude and direction of the weakest electric f

Mars2501 [29]

Answer:

528398.4375 N/C opposite to the direction of the proton

$3.56\times 10^{-8}\ s$

288.24609375 N/C in the same direction of the motion of the electron

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity = $1.8\times 10^{6}\ m/s$

s = Displacement = 3.2 cm

a = Acceleration

Mass of electron = $9.11\times 10^{-31}\ kg$

Mass of electron = $1.67\times 10^{-27}\ kg$

q = Charge of particle = $1.6\times 10^{-19}\ C$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(1.8\times 10^6)^2}{2\times 0.032}\\\Rightarrow a=-5.0625\times 10^{13}\ m/s^2$

Electric field is given by

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{1.67\times 10^{-27}\times -5.0625\times 10^{13}}{1.6\times 10^{-19}}\\\Rightarrow E=−528398.4375\ N/C$

The electric field is 528398.4375 N/C opposite to the direction of the proton

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-1.8\times 10^6}{-5.0625\times 10^{13}}\\\Rightarrow t=3.56\times 10^{-8}\ s$

The time taken is $3.56\times 10^{-8}\ s$

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{9.11\times 10^{-31}\times -5.0625\times 10^{13}}{-1.6\times 10^{-19}}\\\Rightarrow E=288.24609375\ N/C$

The electric field is 288.24609375 N/C in the same direction of the motion of the electron

7 0

4 years ago